How to prove $\int_{-\infty}^{+\infty} f(x)dx = \int_{-\infty}^{+\infty} f\left(x - \frac{1}{x}\right)dx?$ If $f(x)$ is a continuous function on $(-\infty, +\infty)$ and $\int_{-\infty}^{+\infty} f(x) \, dx$ exists. How can I prove that
$$\int_{-\infty}^{+\infty} f(x) \, dx = \int_{-\infty}^{+\infty} f\left( x - \frac{1}{x} \right) \, dx\text{ ?}$$
 A: We can write
\begin{align}
\int_{-\infty}^{\infty}f\left(x-x^{-1}\right)dx&=\int_{0}^{\infty}f\left(x-x^{-1}\right)dx+\int_{-\infty}^{0}f\left(x-x^{-1}\right)dx\\
&=\int_{-\infty}^{\infty}f(2\sinh\theta)\,e^{\theta}d\theta+\int_{-\infty}^{\infty}f(2\sinh\theta)\,e^{-\theta}d\theta\\
&=\int_{-\infty}^{\infty}f(2\sinh\theta)\,2\cosh\theta\,d\theta\\
&=\int_{-\infty}^{\infty}f(x)\,dx.
\end{align}
To pass from the first to the second line, we make the change of variables $x=e^{\theta}$ in the first integral and $x=-e^{-\theta}$ in the second one.
A: Reference:
Glaisher 1876 On a formula of Cauchy's for the evaluation of a class of definite integrals, Proc. Camb. Phil. Soc. III, p. 5-12
This has the original reference to Cauchy, and an added solution by Cayley using a very original change of variable,  "weirder" than those used in previous answers.
A: Here is another way to prove the identity courtesy of @achillehui. Let us generalise the identity using the following lemma.
Lemma :

\begin{align}
\int_{-\infty}^\infty f\left(x\right)\,dx=\int_{-\infty}^\infty f\left(x-\frac{a}{x}\right)\,dx\qquad,\qquad\text{for }\, a>0.
\end{align}

Proof :
Let $\displaystyle\;u(x) = x-\frac{a}{x} $. As $x$ varies over $\mathbb{R}$, we have


*

*$u(x)$ increases monotonically from $-\infty$ at $-\infty$ to $+\infty$ at $0^{-}$.

*$u(x)$ increases monotonically from $-\infty$ at $0^{+}$ to $+\infty$ at $+\infty$.


This means as $x$ varies, $u(x)$ covered $(-\infty,\infty)$ twice.
Let $x_1(u) < 0$ and $x_2(u) > 0$ be the two roots of the equation for a given $u$:
$$u = u(x) =  \frac{x^2-a}{x} \quad\iff\quad x^2 - ux - a = 0\quad\implies\quad x(u)_{1,2}=\frac{1}{2}\left(\;u\pm\sqrt{u^2+4a}\;\right)$$
we have
$$x_1(u) + x_2(u) = u
\quad\implies\quad
\frac{dx_1}{du} + \frac{dx_2}{du} = 1.
$$
From this, we find
$$\begin{align}
\int_{-\infty}^\infty f\left(x-\frac{a}{x}\right)\,dx
&= \left( \int_{-\infty}^{0^{-}} + \int_{0^{+}}^{+\infty}\right) f\left(u(x)\right)\, dx\\
&= \int_{-\infty}^{\infty} f\left(u\right)\,\left(\frac{dx_1}{du} + \frac{dx_2}{du}\right) du\\
&= \int_{-\infty}^{\infty}f\left(u\right)\, du\qquad;\qquad u\mapsto x\\
&= \int_{-\infty}^{\infty}f\left(x\right)\, dx\qquad\qquad\square
\end{align}
$$

Thus, by setting $a=1$, we have

\begin{align}
\int_{-\infty}^\infty f\left(x\right)\,dx=\int_{-\infty}^\infty f\left(x-\frac{1}{x}\right)\,dx
\end{align}

A: Here is yet another way forward.  First, we begin with the integral $I$ given by
$$\begin{align}
I&=\int_{-\infty}^{\infty}f\left(x-\frac1x\right)\,dx \tag 1\\\\
&=\int_{-\infty}^{0}f\left(x-\frac1x\right)\,dx +\int_{0}^{\infty}f\left(x-\frac1x\right)\,dx \tag 2
\end{align}$$
We enforce the substitution $x\to -1/x$ in the integrals on the right-hand side of $(2)$ to obtain
$$\begin{align}
I&=\int_{0}^{\infty}f\left(x-\frac1x\right)\,\left(\frac{1}{x^2}\right)\,dx +\int_{-\infty}^{0}f\left(x-\frac1x\right)\,\left(\frac{1}{x^2}\right)\,dx \tag 3
\end{align}$$
Adding $(2)$ and $(3)$ reveals
$$\begin{align}
2I&=\int_{0}^{\infty}f\left(x-\frac1x\right)\,\left(1+\frac{1}{x^2}\right)\,dx +\int_{-\infty}^{0}f\left(x-\frac1x\right)\,\left(1+\frac{1}{x^2}\right)\,dx \tag 4
\end{align}$$
Next, we enforce the substitution $x-1/x \to x$ in the integrals on the right-hand side of $(4)$ and obtain 
$$\begin{align}
2I&=\int_{-\infty}^{\infty}f\left(x\right)\,dx +\int_{-\infty}^{\infty}f\left(x\right)\,dx \\\\
&=2\int_{-\infty}^{\infty}f\left(x\right)\,dx \tag 5
\end{align}$$
Finally, dividing both sides of $(5)$ by $2$ and using $(1)$ we arrive at
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^{\infty}f\left(x-\frac1x\right)\,dx=\int_{-\infty}^{\infty}f\left(x\right)\,dx }$$
as was to be shown!
A: For $x\ge0$ write the following substitution which maps the positive real domain to the whole of the real domain:
$$ x=1/2\,y+1/2\,\sqrt {{y}^{2}+4}, $$
$${\it dx}= \left( 1/2+1/2\,{\frac {y}{\sqrt {{y}^{2}+4}}} \right) {\it 
dy},$$
and likewise for $x<0$ write the following substitution which maps the negative real domain to the whole of the real domain:$$ x=1/2\,y-1/2\,\sqrt {{y}^{2}+4}, $$
$${\it dx}= \left( 1/2-1/2\,{\frac {y}{\sqrt {{y}^{2}+4}}} \right) {\it 
dy},$$
and in both cases you then have:
$$x-\frac{1}{x}=y,$$ and you then get:
$$\int\limits_{0}^{+\infty} f\left(x-\dfrac{1}{x}\right)dx=\int\limits_{-\infty}^{+\infty} f\left(y\right)\left( 1/2+1/2\,{\frac {y}{\sqrt {{y}^{2}+4}}} \right)dy,$$
$$\int\limits_{-\infty}^{0} f\left(x-\dfrac{1}{x}\right)dx=\int\limits_{-\infty}^{+\infty} f\left(y\right)\left( 1/2-1/2\,{\frac {y}{\sqrt {{y}^{2}+4}}} \right)dy,$$
and therefore:
$$\int\limits_{-\infty}^{+\infty} f\left(x-\dfrac{1}{x}\right)dx=\int\limits_{-\infty}^{+\infty} f\left(y\right)\left( 1/2+1/2\,{\frac {y}{\sqrt {{y}^{2}+4}}} \right)dy+\int\limits_{-\infty}^{+\infty} f\left(y\right)\left( 1/2-1/2\,{\frac {y}{\sqrt {{y}^{2}+4}}} \right)dy,$$
$$=\int\limits_{-\infty}^{+\infty} f\left(y\right)dy$$
A: Here is a neat way of showing the claim without using any "weird" change of variables. Assume $f$ is nice enough to do all that follows. I think $f\in C_c^\infty(R-\{0\})$ is enough. Let
$$F(a) = \int_{-\infty}^\infty f\left(x-\frac{a}{x} \right)dx $$
so that
$$F'(a) = \int_{-\infty}^\infty -\frac{1}{x}f'\left(x-\frac{a}{x} \right)dx.$$
[Please note if there is an error in the following, I've gotten it wrong a couple times.]
Split up the integral and do a change of variables $x=a/y$ so that
$$F'(a)=\int_{\infty}^0 \left[-\frac{y}{a}f'\left(\frac{a}{y}-y\right) \cdot - \frac{a}{y^2} \right]dy+\int_{0}^{-\infty} \left[-\frac{y}{a}f'\left(\frac{a}{y}-y\right) \cdot - \frac{a}{y^2}\right]dy$$
or
$$F'(a)= \int_{-\infty}^\infty -\frac{1}{y}f'\left(\frac{a}{y}-y\right) dy.$$
Here's the tricky part. The original claim is true for odd functions $f$, since both integrals integrate to zero. Thus, only the even part of $f$ matters. We can then assume that $f$ is even, and hence $f'$ is odd. Moving a minus sign out from the argument of $f'$ in the above integral and we get $F'(a)=-F'(a)$. This means that $F'(a)=0$ for all $a$ after appealing to the continuity of $F'(a)$. Hence $F(a)=F(0)$ is constant and 
$$F(1)=\int_{-\infty}^\infty f\left(x-\frac{1}{x} \right)dx = \int_{-\infty}^\infty f\left(x\right)dx=F(0).$$
Approximate any $f$ with a $C_c^\infty$ function supported away from the origin, and the original claim holds.
A: Curiously, I happen to come across this a few days after finding this question, where @achille hui proves the theorem that

Given such a meromorphic function $ϕ(z)$ and any Lebesgue integrable function $f(x)$ on $\mathbb{R}$, we have following identity:
$$∫^∞_{−∞}f(ϕ(x))dx=∫^∞_{−∞}f(x)dx$$

In that question's scope, it is used to show that $$∫^∞_{−∞}f(x-\cot(x))dx=∫^∞_{−∞}f(x)dx$$
Here, $\phi(x)=x-\frac1x$, so to show that
$$∫^∞_{−∞}f\left(x-\frac1x\right)dx=∫^∞_{−∞}f(x)dx$$
we simply have to just check the conditions of the quoted theorem.
We can see that $\phi(x)$ is indeed meromorphic, as its only pole is at the origin and it is holomorphic.
Then, by the quoted theorem, assuming $f(x)$ is Lebesgue integrable over all reals, your case follows as a corollary.
